Question

If α and β are the zeroes of the polynomial p(x) = x2 − 2x +3 find a polynomial whose roots are 1/ α and 1/ β

Answer 1

Step-by-step explanation: P(X) = X^2 - 2X+3 P(X) = X^2 - 3X - X + 3 P(X) = X(X - 3) +1 (X - 3 ) P(X) = (X +1) ( X -3) ROOTS ARE -1 ,3 FOR THE EQUATION 1/a =1 , a =. - 1 1/b = 3 ,b = 1/3we know that form of qudratic equation is x^2 -( a+ b ) x + (a ×b) x^2 -(-1+1/3)x +(-1×1/3) x^2 - 2/3x-1/3 multiply both sides with 3 3x^2 - 2x - 1 THIS IS THE CORRECT ANSWER

Answer 2

★ Correct question -:

$\alpha \: and \beta \: are \: roots \: of \: p(x)$

p(x) = x² - 2x - 3

★ To find :

The polynomial whose roots are $\frac{1}{\alpha} \: and \: \frac{1}{\beta}$

★ Solution :

x² - 2x - 3

by middle term splitting

x² - 3x + x - 3

x(x - 3) + 1 (x-3)

(x + 1) (x - 3)

→ therefore roots of p(x) are -1 , 3 .

✏️ To find polynomial whose roots are 1/alpha and 1/beta

→ $\alpha$ = -1

→ $\beta$ = 3

★ The roots of unknown polynomial are -:

→ $\frac{1}{-1} = -1$

→ $\frac{1}{3}$

✏️ it has two roots I.e it is a quadratic polynomial..

We know that,

✏️ Form of quadratic equation =

$x^2 - (\alpha + \beta)x + (\alpha \times \beta)$

✏️ by putting the values of alpha and beta...

$x^2 - \left( -1 + \frac{1}{3}\right) x + \left( -1 \times \frac{1}{3}\right)$

$\implies x^2 - \left(\frac{-2}{3}\right) x - \frac{1}{3}$

$\purple{\leadsto x^2 + \frac{2}{3}x - \frac{1}{3}}$

✏️ To make it a polynomial :

Multiply by 3 to make whole numbers as coefficients ..

$3 \times \left(x^2 + \frac{2}{3}x - \frac{1}{3}\right)$

$\purple{\leadsto 3x^2 + 2x - 1 }$